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Stiffness, strength and related topics
StiiFness, strength and related topics
6.1 I N T R O D U C T I O N Often we need to ensure that a component does not break, i.e., it has adequate strength, or that it does not suffer excessive deformation under load, i.e., it has adequate stiffness. F o r these reasons we shall discuss methods of calculating the stiffness and strength of composites in the following chapters. In particular we shall be concerned with the behaviour of flat, plate-like elements, as composite components are frequently fab ricated in this way. Since most of these methods are derived from the basic principles of mechanics and stress analysis, the latter are reviewed below before going on t o consider composites. Full details of these methods can be found in standard texts on such topics.
6.2 L O A D S A N D D E F O R M A T I O N S 6.2.1 Loads We consider the loads acting on a structure or component to be of two basic types; forces and couples. Forces can be further subdivided into tensile, compressive
and shear,
a n d couples into bending
moments
and
twisting
moments (or torques). All of these can act in isolation or in combination, depending on the situation. 6.2.2 Structural entities When carrying out a structural analysis it is convenient to represent the real structure by one, or more, idealized elements. Each element will almost
STRESS AND STRAIN certainly be a simplification of the actual situation, in that we will make assumptions about the way the loads are distributed within it. These simplifications are often based on the dimensions of the part. A straight member whose length is much greater that its cross-sectional dimensions, and is subjected to a tension or compression along its length, is referred to as a rod. A similar member, but subjected to a bending moment is known as a beam, and when loaded by a torque is called a shaft. When we have an element whose length and breadth are much greater than its thickness, and which is subjected to in-plane forces, we refer to it as a membrane. If we subject it to bending moments we call it a plate. Cylindrical components can be treated as membranes provided the diameter is constant and much larger than the wall thickness. If the diameter of the cylinder varies along its length we would call it a shell; in this case its behaviour is similar to, but more complicated than, a plate.
6.2.3 Deformations Each type of load referred to above produces a characteristic deformation of the body on which it acts. These are shown in Figure 6.1. We also frequently refer to states of plane stress and plane strain. In plane stress, stresses normal to the plane in which the stresses act are considered to be zero, e.g. a plate in simple bending. In plane strain, strains normal to the plane in which the stresses act are considered to be zero, e.g. the crosssection of a bar held between rigid walls.
6.3 STRESS A N D S T R A I N 6.3.1 Stress To allow us to deal with components of differing size we work with normalized loads. Thus tensile and compressive forces are divided by cross-sectional area to give the corresponding direct stress (σ) and shear force by a surface area to give a shear stress (τ) (see Figure 6.2). Stress therefore has units of force/area, i.e., N/m^ or Pascal (Pa). A bending moment alone will give rise to a direct stress which varies Hnearly through the depth of the component from tensile on one surface, to compressive on the other; a transverse force (normal to the longitudinal axis of the beam) will cause shear stresses as well as direct stresses (see Figures 6.2 and 2.24). A torque will give rise to shear stresses. Again, these will vary in a linear fashion but can only be easily calculated for simple compon ents (such as a circular section rod - see Figure 6.2). For plates it is usual to work with loads (forces and couples) per unit width, as shown in Figure 6.3.
207
Deformation
Load
I
I
Extension
Tensile force
\ Final position
Contraction
Compressive force
Shear force
Bending moment
Torque
I
Figure 6.1
Loads
Twisting
and associated deformations.
A cross-section area of rod
Direct stress (tensile or compressive) σ=Ρ/4
area A
Shear stress τ = WjA
Tension
Direct stress
o=My/f
I is 2nd moment of area of beam cross-section
Compression
Shear stress τ = Tr/o
J is polar 2nd moment of area of shaft cross-section Figure 6.2
Loads and associated stresses.
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STIFFNESS, STRENGTH A N D RELATED TOPICS
Figure 6.3 Loads acting on a plate: Force per unit width, N = P/w; moment per unit width, Μ = Mo/w; direct stress due to P, σ = N/t=^P/wt.
6.3.2 Strain Strain is a non-dimensional measure of deformation; it can be elastic (instantaneous, reversible), anelastic (reversible, time dependent), inelastic (irreversible, time dependent) or permanent. There are two types of strain, direct and shear, corresponding to the appropriate stress. Definitions (socalled 'engineering strain') are obtained from the quantities given in Figure 6.1. So, direct strain ε = δ/1 taken as positive for a tensile force and negative for compression; shear strain Ύ = φι +2· For bending, the radius of curvature is related to the direct strain by
k being the curvature. The deflection can be related to the curvature using standard mathematical methods. For the simple case of twisting shown in Figure 6.2, shear strain is given by y = re/l 6.4 S T R E S S - S T R A I N R E L A T I O N S When solving stress analysis problems, we usually know the stresses (from the applied loads) and wish to calculate the corresponding strains (or deformations). For small (i.e., elastic) strains, these quantities are related by the elastic constants which will be known for the material under consider ation. For the simple case of one-dimensional loading, direct stress and strain are related by Young's modulus, E, i.e. σ-Εε
or
ε = σ/Ε,
(6.1)
211
STRESS-STRAIN RELATIONS
and shear stress and strain are related by the shear modulus, G, i.e. x = Gy
or
y = x/G.
(6.2)
Any longitudinal deformation (in the direction of the force) will be accom panied by a lateral deformation in the opposite sense, as illustrated for a plate in Figure 6.4.
r
1
1 1
1 1 1 1 I
1
y
X
1
1
1
1 1
Longitudinal tensile force; lateral contraction
Figure 6.4
Longitudinal compressive force; lateral expansion
Effect of Poisson's ratio.
The strains in the two orthogonal directions are related by a quantity known as Poisson's ratio, v, i.e.
It follows, then, that if direct stresses are acting in two orthogonal direc tions, the resultant strain in each direction will be dependent on both stresses, i.e. (6.3) The above equations refer to a material for which stress is proportional to strain, i.e., the material is linear elastic. Such materials are said to obey Hooke's Law. Linear behaviour is not always seen, especially for shear loading. Usually we also need stresses in terms of strain. These are easily found by solving equations 6.3, i.e.
(6.4)
212
STIFFNESS, STRENGTH A N D RELATED TOPICS
6.5 B E N D I N G O F P L A T E S The influence of Poisson's ratio is seen, also, in the bending of plates. As already stated, bending will result in a hnear distribution of direct stress varying through the thickness, from tension on one surface to compression on the other. Associated with the tensile stress we see now that there will be a lateral contraction, and associated with the compressive stress a lateral expansion. The result of this is to bend the plate in the opposite sense to the main bending (Figure 6.5), giving what is known as anticlastic curvature.
Anticlastic curvature (ky) arising from Poisson's ratio
Applied bending moment
Primary, ^ applied, curvature (/rj
Figure 6.5 Bending deformation of a plate. Tensile stress on upper surface pro duces lateral contraction (opposite on the lower surface).
Using a generalization of beam bending (section 6.3) we can relate the direct strains and curvatures arising from applied bending moments, as ex = z/Rx = zkx,
and
Ey = Z/Ry = Zky,
ζ being the distance measured through the thickness. For an applied twisting moment there will be a corresponding shear strain given by yxy = zk^r
Note the use of a double subscript when dealing with shear. The first subscript relates to the normal of the plane on which the stress acts and the second subscript to the direction of the stress (see also Figure 6.7). If we substitute the above expressions for direct strain into equations 6.3 we can show that the corresponding direct stresses are (T, = ( £ z / ( l - v 2 ) ) ( / c , + vfc,),
ay = (Ez/(\-v'))ivk,
+ ky).
(6.5)
213
ISOTROPIC MATERIALS
Methods for finding deflections are given in advanced texts on applied mechanics.
6.6 I S O T R O P I C M A T E R I A L S The equations used above relate to isotropic materials, such as unre inforced metals or polymers in certain conditions. In these materials the elastic constrants, £, G and v, do not vary with direction. In other words, if we were to take a sheet of metal with random orientation of the grains, say, and cut from it a series of test specimens, as illustrated in Figure 6.6, the value of Ε determined from a tensile test would be the same for each specimen.
Figure 6.6
Test specimens taken from an isotropic sheet.
A further property of isotropic materials is that only two elastic constants are needed to describe (or characterize) the stress-strain behaviour. Usually we use Ε and v, which are related to G by the equation G =
Example
2(1+v)
(6.6)
6.1
A plate is subjected to the set of stresses shown in the figure below. Note that the shear stress is shown here as positive. Determine the associated strains if the material is isotropic with £ = 70 G P a and v = 0.3.
214
STIFFNESS, STRENGTH AND RELATED TOPICS 300 MPa
150 MPa
500 MPa
Let σχ = 500,
(Ty = - 300 (compressive).
and τ , , = 150.
Using equations 6.3 5 0 0 x 10^
0 . 3 ( - 3 0 0 x 10^)
70x10^
70x10^
- 0 . 3 x 5 0 0 x 10^ 7 0 x 1 0^
= 8.43x10"^ (tensile).
( - 3 0 0 x 10^) 70x10^ "
"^'"^^
^
(compressive).
Using equations 6.2 and 6.6, γ = T^y/G = 2τ^3,(1 + ν)/Ε = 5.57 χ ΙΟ" ^
6.7 P R I N C I P A L STRESS A N D S T R A I N When determining the strength of an isotropic material it is conventional to use principal stresses or strains in conjunction with a failure criterion. Principal values are the maxima at a point in a component and are found from the stresses, or strains, at that point, expressed in a convenient set of orthogonal axes. When the stresses vary throughout the component, as is usually the case, it is necessary to calculate the principal values at several points to find the absolute maximum. In order to find the principal stresses at a point, it is first necessary to obtain an expression for the direct and shear stresses acting on a face at some angle θ to the basic axes as shown in Figure 6.7. N o t e that all the
215
PRINCIPAL STRESS A N D STRAIN
(a)
(b) Figure 6.7 (a) Element in n-m axes oriented at θ to element in x-y axes; (b) stresses on faces of triangles A and Β used to related stresses in the two axes systems.
stresses are shown as acting in a positive sense. The relevant set of transformation equations is σ„ = σχ cos^ O-hCy sin^ Θ+Τχγΐ = Tnm =
sin^ Ö -h Oy cos^ O-x^yl —
sin
Ö cos
0 -f Oy sin
sin Θ cos Θ, sin θ cos
Ö cos
Ö 4· T , ^ ( C O S ^ Ö — sin^
Θ).
(6.7)
By differentiating equation 6.7 with respect to θ the maximum, i.e., principal, values of the direct stresses can be shown to be σ.,, = ^ ± ^ [ ( < τ . - σ , ) '
+ 4τ^]"^
(6.8a)
STIFFNESS, STRENGTH AND RELATED TOPICS
216
and the maximum shear stress (6.8b) We see, then, that application of only a shear stress (τ^^^) will give rise to (principal) direct stresses and, conversely, application of only direct stresses {σ^ and σ^) will give rise to a (maximum) shear stress. Similar transformation equations can be found for strains. The latter can be obtained from the former if we replace direct stress by direct strain and shear stress by (shear strain)/2, i.e. Ex = Ex
=
hxy =-^x
cos^
e-\-Ey
sin^ 0-h2 7xy2sin ö c o s ö
sin^ 0 + ε^, cos^ O-^y^yl
sin θ cos θ
sin Ö cos θ -f ε^ sin θ cos Θ + iTxyicos^ θ - sin^ Θ)
(6.9)
For isotropic materials the directions of principal stresses and strains coincide. It is often convenient to write the transformation equations 6.7 and 6.9 in a 'shorthand' form using matrix notation (see Appendix for a summary of matrix algebra). Adopting this approach we get ffnm
(6.10)
= Tff^^, and
(6.11) where
^xy = {^x
hiyxy}.
and the transformation matrix Imn
T=
(6.12)
— Imn \_ — mn
mn
(m^ —
n^)j
where m = cos Ö, ^2 = sin Θ. We see that the same transformation can be employed for both stress and strain provided we use yjl in the strain matrix, i.e. we use ε instead of ε.
217
THIN-WALLED CYLINDERS A N D SPHERES
6.8 T H I N - W A L L E D C Y L I N D E R S A N D S P H E R E S A frequently-encountered situation, in which a two-dimensional set of stresses exists, is that of a thin-walled cylinder or sphere subjected to internal pressure, p . 6.8.1 Cylinder with closed ends Figure 6.8 shows the stresses in the axial direction (σ^) and hoop, or circumferential, direction (σπ) which are given by o^ = pR/2u
and
^h = pRA,
(6.13)
i, being the wall thickness and R the tube radius measured to the mid-line of the wall. Following equation 6.3 we can obtain the corresponding strains
(6.14) If we substitute from equation 6.13 into equation 6.14 we get £iV2 (6.15)
Figure 6.8
Stresses in wall of pressurized cylinder.
6.8.2 Sphere From the essential symmetry of the sphere seen in Figure 6.9 we see that orthogonal stresses, in any set of axes, must be the same. It can be shown that the stress for a sphere of radius R and wall thickness t is a, = pR/2t.
(6.16)
STIFFNESS, STRENGTH AND RELATED TOPICS
218
Figure 6.9 Stress in wall of pressurized sphere. Example
6,2
A closed pipe 2 m long with 500 mm diameter and 20 m m wall thickness is subjected to an internal pressure of 0.5 M P a . Calculate the change of length and diameter caused by the pressure. Take Young's modulus to be 1 G P a and Poisson's ratio as 0.45. F r o m equation 6.15 pR^,
,
0.5x10^x250,^, 1x10^x20 =0.3125x10
^
^'-'-'-''^
-3
and
Hence the change in length (L) =€aL = 0.3125 X 10"^ X 2 m = 0.615 mm. Now, the change in diameter (D) is related to the change in circumfer ence (C), i.e. C = nD (C•^-δC) = π{D + δD)
That is. ^ ÖD ÖD l + e c = l + — , or £ ο = —
but
Be
= €h,
so
5 D = επ D.
Hence the change in diameter = 4.838 χ 10"^ χ 500 = 2.419 mm.
219
SUMMARY
6.9 F A I L U R E CRITERIA Having determined the principal stress or strain we would use these maxima in an appropriate failure criterion to predict whether, or not, our material will fail, failure being taken as yielding, i.e., onset of permanent deformation, or fracture. For isotropic materials there are a large number of criteria, the appropri ate choice depending on the material (is it brittle or ductile?) and the local stress field (one-, two- or three-dimensional). There is no single criterion that is universally applicable. A simple criterion is the maximum principal stress theory which states that failure will occur, in a material under a multi-axial stress field, when the maximum principal stress attains a value equal to the yield stress (σ^) obtained from a uniaxial tensile test, i.e. (6.17)
Gi=ay.
A more complicated criterion, often known as the von Mises criterion is derived from considerations of shear strain energy. The equation defining failure for a three-dimensional stress state is (σ 1 - σ2)' + (σ2 -σ,f-l·{σ,-σ,f
= 1σ]
(6.18)
where σ ι , σ2 and σ3 are the principal direct stresses. Under plane stress conditions, such as found in a thin sheet subjected to biaxial loading, equation 6.18 reduces to σ? + σ^-σισ2 = σ2.
(6.19)
6.10 S U M M A R Y In this chapter we have reviewed the fundamental parameters and relation ships associated with the stress analysis of isotropic materials. The solution of the relevant equations enables us to find stresses and strains (and therefore deformations) and hence assess whether a material will fail under a defined set of loads. We shall see in later chapters that similar equations appear in the analysis of composites, differing only in detail from those used for isotropic materials. In other words, n o new concepts are needed.
FURTHER READING General Benham, P. P. and Warnock, F. V. (1982) Mechanics of Solids and Structures, Pitman. Crandall, S. H., Dahl, Ν. C. and Lardner, Τ J. (1978) An Introduction to the Mechanics of Solids, McGraw Hill.
STIFFNESS, STRENGTH AND RELATED TOPICS
220
PROBLEMS 6.1 A sphere has a diameter of 250 m m and a wall thickness of 10 mm. The material from which the sphere is fabricated is isotropic with Young's modulus of 70 G P a and Poisson's ratio of 0.33. Calculate the change in the diameter of the sphere when it is subjected to an internal pressure of 0.75 M P a . 6.2 A thin sheet of plastic is loaded by a two-dimensional set of stresses: (T^ = 40, σ^ = 20, T;cy = 2 5 M P a . The elastic properties of the plastic are £ = 3 G P a and ν = 0.35. The yield stress obtained from a simple tensile test is 56 M P a . Determine whether the sheet has failed according to (a) the maximum principal stress theory and (b) the von Mises criterion.
SELF-ASSESSMENT Q U E S T I O N S Indicate whether statements 1 to 8 are true or false. 1. A structural member whose length is much greater than its cross-section dimensions is always referred to as a rod. (A) true (B) false 2. The thickness of a plate is of similar magnitude to its length and breadth. (A) true (B) false 3. Cylindrical components can be treated as membranes provided the diameter is much larger than the wall thickness. (A) true (B) false 4. There are three types of stress; direct, shear and bending stress. (A) true (B) false 5. A plate subjected to a unidirectional direct stress experiences strain only in the direction of the stress. (A) true (B) false
SUMMARY 6. The elastic stress-strain behaviour of an isotropic material can be characterized by only two elastic constants. (A) true (B) false 7. Principal stresses are the maximum and minimum shear stresses at a point in a body. (A) true (B) false 8. The failure of a stressed component can always be predicted by the von Mises criterion. (A) true (B) false For each of the statements of questions 9 to 11, one or more of the completions given are correct. Mark the correct completions. 9. Stress (A) (B) (C) (D) (E)
is the total load on a component, is a normalized measure of load, arises only from the application of a couple, always exists as two components at right angles, has units of force/length.
10. Strain (A) (B) (C) (D) (E)
is an alternative term for displacement, has units of 1/length, is a non-dimensional measure of deformation, is caused by the application of a stress, can be related to curvature of a bent beam.
11. Poisson's ratio (A) (B) (C) (D)
defines the curvature of a plate, is an alternative property to shear modulus, has the same units as stress, relates a longitudinal direct strain to the corresponding transverse strain, (E) is the same as shear strain. Each of the sentences in questions 12 to 16 consists of an assertion followed by a reason. Answer: (A) if both assertion and reason are true statements and the reason is a correct explanation of the assertion,
221
STIFFNESS, STRENGTH AND RELATED TOPICS
222 (B) if a (C) if (D) if (E) if
both assertion and reason are true statements but the reason is not correct explanation of the assertion, the assertion is true but the reason is a false statement, the assertion is false but the reason is a true statement, both the assertion and reason are false statement.
12. The loads acting on a component are sub-divided into forces and couples because forces can be sub-divided into tension and compression. 13. In a plate loaded in its plane the orthogonal direct strains are deter mined by both direct stresses because the shear strain is determined by the shear stress. 14. An isotropic material always behaves in a hnear elastic fashion shear modulus is related to Young's modulus and Poisson's ratio.
because
15. A plate loaded by couples will exhibit anticlastic curvature because direct stresses vary linearly through the thickness.
the
16. In a pressurized cylinder the axial and h o o p stresses have the same magnitude because the axial strain is greater than the circumferential strain.
ANSWERS Problems 6.1 6.2
0,011mm, (a) Yes, (b) No.
Self-assessment L B ; 2, B; 3. A; 4. B; 5, B; 6. A; 7. B; 8, B; 9. B; 10, C, D, E; 11. D; 12. B; 13, B; 14. D; 15. B; 16. E.
6.1 I N T R O D U C T I O N Often we need to ensure that a component does not break, i.e., it has adequate strength, or that it does not suffer excessive deformation under load, i.e., it has adequate stiffness. F o r these reasons we shall discuss methods of calculating the stiffness and strength of composites in the following chapters. In particular we shall be concerned with the behaviour of flat, plate-like elements, as composite components are frequently fab ricated in this way. Since most of these methods are derived from the basic principles of mechanics and stress analysis, the latter are reviewed below before going on t o consider composites. Full details of these methods can be found in standard texts on such topics.
6.2 L O A D S A N D D E F O R M A T I O N S 6.2.1 Loads We consider the loads acting on a structure or component to be of two basic types; forces and couples. Forces can be further subdivided into tensile, compressive
and shear,
a n d couples into bending
moments
and
twisting
moments (or torques). All of these can act in isolation or in combination, depending on the situation. 6.2.2 Structural entities When carrying out a structural analysis it is convenient to represent the real structure by one, or more, idealized elements. Each element will almost
STRESS AND STRAIN certainly be a simplification of the actual situation, in that we will make assumptions about the way the loads are distributed within it. These simplifications are often based on the dimensions of the part. A straight member whose length is much greater that its cross-sectional dimensions, and is subjected to a tension or compression along its length, is referred to as a rod. A similar member, but subjected to a bending moment is known as a beam, and when loaded by a torque is called a shaft. When we have an element whose length and breadth are much greater than its thickness, and which is subjected to in-plane forces, we refer to it as a membrane. If we subject it to bending moments we call it a plate. Cylindrical components can be treated as membranes provided the diameter is constant and much larger than the wall thickness. If the diameter of the cylinder varies along its length we would call it a shell; in this case its behaviour is similar to, but more complicated than, a plate.
6.2.3 Deformations Each type of load referred to above produces a characteristic deformation of the body on which it acts. These are shown in Figure 6.1. We also frequently refer to states of plane stress and plane strain. In plane stress, stresses normal to the plane in which the stresses act are considered to be zero, e.g. a plate in simple bending. In plane strain, strains normal to the plane in which the stresses act are considered to be zero, e.g. the crosssection of a bar held between rigid walls.
6.3 STRESS A N D S T R A I N 6.3.1 Stress To allow us to deal with components of differing size we work with normalized loads. Thus tensile and compressive forces are divided by cross-sectional area to give the corresponding direct stress (σ) and shear force by a surface area to give a shear stress (τ) (see Figure 6.2). Stress therefore has units of force/area, i.e., N/m^ or Pascal (Pa). A bending moment alone will give rise to a direct stress which varies Hnearly through the depth of the component from tensile on one surface, to compressive on the other; a transverse force (normal to the longitudinal axis of the beam) will cause shear stresses as well as direct stresses (see Figures 6.2 and 2.24). A torque will give rise to shear stresses. Again, these will vary in a linear fashion but can only be easily calculated for simple compon ents (such as a circular section rod - see Figure 6.2). For plates it is usual to work with loads (forces and couples) per unit width, as shown in Figure 6.3.
207
Deformation
Load
I
I
Extension
Tensile force
\ Final position
Contraction
Compressive force
Shear force
Bending moment
Torque
I
Figure 6.1
Loads
Twisting
and associated deformations.
A cross-section area of rod
Direct stress (tensile or compressive) σ=Ρ/4
area A
Shear stress τ = WjA
Tension
Direct stress
o=My/f
I is 2nd moment of area of beam cross-section
Compression
Shear stress τ = Tr/o
J is polar 2nd moment of area of shaft cross-section Figure 6.2
Loads and associated stresses.
210
STIFFNESS, STRENGTH A N D RELATED TOPICS
Figure 6.3 Loads acting on a plate: Force per unit width, N = P/w; moment per unit width, Μ = Mo/w; direct stress due to P, σ = N/t=^P/wt.
6.3.2 Strain Strain is a non-dimensional measure of deformation; it can be elastic (instantaneous, reversible), anelastic (reversible, time dependent), inelastic (irreversible, time dependent) or permanent. There are two types of strain, direct and shear, corresponding to the appropriate stress. Definitions (socalled 'engineering strain') are obtained from the quantities given in Figure 6.1. So, direct strain ε = δ/1 taken as positive for a tensile force and negative for compression; shear strain Ύ = φι +2· For bending, the radius of curvature is related to the direct strain by
k being the curvature. The deflection can be related to the curvature using standard mathematical methods. For the simple case of twisting shown in Figure 6.2, shear strain is given by y = re/l 6.4 S T R E S S - S T R A I N R E L A T I O N S When solving stress analysis problems, we usually know the stresses (from the applied loads) and wish to calculate the corresponding strains (or deformations). For small (i.e., elastic) strains, these quantities are related by the elastic constants which will be known for the material under consider ation. For the simple case of one-dimensional loading, direct stress and strain are related by Young's modulus, E, i.e. σ-Εε
or
ε = σ/Ε,
(6.1)
211
STRESS-STRAIN RELATIONS
and shear stress and strain are related by the shear modulus, G, i.e. x = Gy
or
y = x/G.
(6.2)
Any longitudinal deformation (in the direction of the force) will be accom panied by a lateral deformation in the opposite sense, as illustrated for a plate in Figure 6.4.
r
1
1 1
1 1 1 1 I
1
y
X
1
1
1
1 1
Longitudinal tensile force; lateral contraction
Figure 6.4
Longitudinal compressive force; lateral expansion
Effect of Poisson's ratio.
The strains in the two orthogonal directions are related by a quantity known as Poisson's ratio, v, i.e.
It follows, then, that if direct stresses are acting in two orthogonal direc tions, the resultant strain in each direction will be dependent on both stresses, i.e. (6.3) The above equations refer to a material for which stress is proportional to strain, i.e., the material is linear elastic. Such materials are said to obey Hooke's Law. Linear behaviour is not always seen, especially for shear loading. Usually we also need stresses in terms of strain. These are easily found by solving equations 6.3, i.e.
(6.4)
212
STIFFNESS, STRENGTH A N D RELATED TOPICS
6.5 B E N D I N G O F P L A T E S The influence of Poisson's ratio is seen, also, in the bending of plates. As already stated, bending will result in a hnear distribution of direct stress varying through the thickness, from tension on one surface to compression on the other. Associated with the tensile stress we see now that there will be a lateral contraction, and associated with the compressive stress a lateral expansion. The result of this is to bend the plate in the opposite sense to the main bending (Figure 6.5), giving what is known as anticlastic curvature.
Anticlastic curvature (ky) arising from Poisson's ratio
Applied bending moment
Primary, ^ applied, curvature (/rj
Figure 6.5 Bending deformation of a plate. Tensile stress on upper surface pro duces lateral contraction (opposite on the lower surface).
Using a generalization of beam bending (section 6.3) we can relate the direct strains and curvatures arising from applied bending moments, as ex = z/Rx = zkx,
and
Ey = Z/Ry = Zky,
ζ being the distance measured through the thickness. For an applied twisting moment there will be a corresponding shear strain given by yxy = zk^r
Note the use of a double subscript when dealing with shear. The first subscript relates to the normal of the plane on which the stress acts and the second subscript to the direction of the stress (see also Figure 6.7). If we substitute the above expressions for direct strain into equations 6.3 we can show that the corresponding direct stresses are (T, = ( £ z / ( l - v 2 ) ) ( / c , + vfc,),
ay = (Ez/(\-v'))ivk,
+ ky).
(6.5)
213
ISOTROPIC MATERIALS
Methods for finding deflections are given in advanced texts on applied mechanics.
6.6 I S O T R O P I C M A T E R I A L S The equations used above relate to isotropic materials, such as unre inforced metals or polymers in certain conditions. In these materials the elastic constrants, £, G and v, do not vary with direction. In other words, if we were to take a sheet of metal with random orientation of the grains, say, and cut from it a series of test specimens, as illustrated in Figure 6.6, the value of Ε determined from a tensile test would be the same for each specimen.
Figure 6.6
Test specimens taken from an isotropic sheet.
A further property of isotropic materials is that only two elastic constants are needed to describe (or characterize) the stress-strain behaviour. Usually we use Ε and v, which are related to G by the equation G =
Example
2(1+v)
(6.6)
6.1
A plate is subjected to the set of stresses shown in the figure below. Note that the shear stress is shown here as positive. Determine the associated strains if the material is isotropic with £ = 70 G P a and v = 0.3.
214
STIFFNESS, STRENGTH AND RELATED TOPICS 300 MPa
150 MPa
500 MPa
Let σχ = 500,
(Ty = - 300 (compressive).
and τ , , = 150.
Using equations 6.3 5 0 0 x 10^
0 . 3 ( - 3 0 0 x 10^)
70x10^
70x10^
- 0 . 3 x 5 0 0 x 10^ 7 0 x 1 0^
= 8.43x10"^ (tensile).
( - 3 0 0 x 10^) 70x10^ "
"^'"^^
^
(compressive).
Using equations 6.2 and 6.6, γ = T^y/G = 2τ^3,(1 + ν)/Ε = 5.57 χ ΙΟ" ^
6.7 P R I N C I P A L STRESS A N D S T R A I N When determining the strength of an isotropic material it is conventional to use principal stresses or strains in conjunction with a failure criterion. Principal values are the maxima at a point in a component and are found from the stresses, or strains, at that point, expressed in a convenient set of orthogonal axes. When the stresses vary throughout the component, as is usually the case, it is necessary to calculate the principal values at several points to find the absolute maximum. In order to find the principal stresses at a point, it is first necessary to obtain an expression for the direct and shear stresses acting on a face at some angle θ to the basic axes as shown in Figure 6.7. N o t e that all the
215
PRINCIPAL STRESS A N D STRAIN
(a)
(b) Figure 6.7 (a) Element in n-m axes oriented at θ to element in x-y axes; (b) stresses on faces of triangles A and Β used to related stresses in the two axes systems.
stresses are shown as acting in a positive sense. The relevant set of transformation equations is σ„ = σχ cos^ O-hCy sin^ Θ+Τχγΐ = Tnm =
sin^ Ö -h Oy cos^ O-x^yl —
sin
Ö cos
0 -f Oy sin
sin Θ cos Θ, sin θ cos
Ö cos
Ö 4· T , ^ ( C O S ^ Ö — sin^
Θ).
(6.7)
By differentiating equation 6.7 with respect to θ the maximum, i.e., principal, values of the direct stresses can be shown to be σ.,, = ^ ± ^ [ ( < τ . - σ , ) '
+ 4τ^]"^
(6.8a)
STIFFNESS, STRENGTH AND RELATED TOPICS
216
and the maximum shear stress (6.8b) We see, then, that application of only a shear stress (τ^^^) will give rise to (principal) direct stresses and, conversely, application of only direct stresses {σ^ and σ^) will give rise to a (maximum) shear stress. Similar transformation equations can be found for strains. The latter can be obtained from the former if we replace direct stress by direct strain and shear stress by (shear strain)/2, i.e. Ex = Ex
=
hxy =-^x
cos^
e-\-Ey
sin^ 0-h2 7xy2sin ö c o s ö
sin^ 0 + ε^, cos^ O-^y^yl
sin θ cos θ
sin Ö cos θ -f ε^ sin θ cos Θ + iTxyicos^ θ - sin^ Θ)
(6.9)
For isotropic materials the directions of principal stresses and strains coincide. It is often convenient to write the transformation equations 6.7 and 6.9 in a 'shorthand' form using matrix notation (see Appendix for a summary of matrix algebra). Adopting this approach we get ffnm
(6.10)
= Tff^^, and
(6.11) where
^xy = {^x
hiyxy}.
and the transformation matrix Imn
T=
(6.12)
— Imn \_ — mn
mn
(m^ —
n^)j
where m = cos Ö, ^2 = sin Θ. We see that the same transformation can be employed for both stress and strain provided we use yjl in the strain matrix, i.e. we use ε instead of ε.
217
THIN-WALLED CYLINDERS A N D SPHERES
6.8 T H I N - W A L L E D C Y L I N D E R S A N D S P H E R E S A frequently-encountered situation, in which a two-dimensional set of stresses exists, is that of a thin-walled cylinder or sphere subjected to internal pressure, p . 6.8.1 Cylinder with closed ends Figure 6.8 shows the stresses in the axial direction (σ^) and hoop, or circumferential, direction (σπ) which are given by o^ = pR/2u
and
^h = pRA,
(6.13)
i, being the wall thickness and R the tube radius measured to the mid-line of the wall. Following equation 6.3 we can obtain the corresponding strains
(6.14) If we substitute from equation 6.13 into equation 6.14 we get £iV2 (6.15)
Figure 6.8
Stresses in wall of pressurized cylinder.
6.8.2 Sphere From the essential symmetry of the sphere seen in Figure 6.9 we see that orthogonal stresses, in any set of axes, must be the same. It can be shown that the stress for a sphere of radius R and wall thickness t is a, = pR/2t.
(6.16)
STIFFNESS, STRENGTH AND RELATED TOPICS
218
Figure 6.9 Stress in wall of pressurized sphere. Example
6,2
A closed pipe 2 m long with 500 mm diameter and 20 m m wall thickness is subjected to an internal pressure of 0.5 M P a . Calculate the change of length and diameter caused by the pressure. Take Young's modulus to be 1 G P a and Poisson's ratio as 0.45. F r o m equation 6.15 pR^,
,
0.5x10^x250,^, 1x10^x20 =0.3125x10
^
^'-'-'-''^
-3
and
Hence the change in length (L) =€aL = 0.3125 X 10"^ X 2 m = 0.615 mm. Now, the change in diameter (D) is related to the change in circumfer ence (C), i.e. C = nD (C•^-δC) = π{D + δD)
That is. ^ ÖD ÖD l + e c = l + — , or £ ο = —
but
Be
= €h,
so
5 D = επ D.
Hence the change in diameter = 4.838 χ 10"^ χ 500 = 2.419 mm.
219
SUMMARY
6.9 F A I L U R E CRITERIA Having determined the principal stress or strain we would use these maxima in an appropriate failure criterion to predict whether, or not, our material will fail, failure being taken as yielding, i.e., onset of permanent deformation, or fracture. For isotropic materials there are a large number of criteria, the appropri ate choice depending on the material (is it brittle or ductile?) and the local stress field (one-, two- or three-dimensional). There is no single criterion that is universally applicable. A simple criterion is the maximum principal stress theory which states that failure will occur, in a material under a multi-axial stress field, when the maximum principal stress attains a value equal to the yield stress (σ^) obtained from a uniaxial tensile test, i.e. (6.17)
Gi=ay.
A more complicated criterion, often known as the von Mises criterion is derived from considerations of shear strain energy. The equation defining failure for a three-dimensional stress state is (σ 1 - σ2)' + (σ2 -σ,f-l·{σ,-σ,f
= 1σ]
(6.18)
where σ ι , σ2 and σ3 are the principal direct stresses. Under plane stress conditions, such as found in a thin sheet subjected to biaxial loading, equation 6.18 reduces to σ? + σ^-σισ2 = σ2.
(6.19)
6.10 S U M M A R Y In this chapter we have reviewed the fundamental parameters and relation ships associated with the stress analysis of isotropic materials. The solution of the relevant equations enables us to find stresses and strains (and therefore deformations) and hence assess whether a material will fail under a defined set of loads. We shall see in later chapters that similar equations appear in the analysis of composites, differing only in detail from those used for isotropic materials. In other words, n o new concepts are needed.
FURTHER READING General Benham, P. P. and Warnock, F. V. (1982) Mechanics of Solids and Structures, Pitman. Crandall, S. H., Dahl, Ν. C. and Lardner, Τ J. (1978) An Introduction to the Mechanics of Solids, McGraw Hill.
STIFFNESS, STRENGTH AND RELATED TOPICS
220
PROBLEMS 6.1 A sphere has a diameter of 250 m m and a wall thickness of 10 mm. The material from which the sphere is fabricated is isotropic with Young's modulus of 70 G P a and Poisson's ratio of 0.33. Calculate the change in the diameter of the sphere when it is subjected to an internal pressure of 0.75 M P a . 6.2 A thin sheet of plastic is loaded by a two-dimensional set of stresses: (T^ = 40, σ^ = 20, T;cy = 2 5 M P a . The elastic properties of the plastic are £ = 3 G P a and ν = 0.35. The yield stress obtained from a simple tensile test is 56 M P a . Determine whether the sheet has failed according to (a) the maximum principal stress theory and (b) the von Mises criterion.
SELF-ASSESSMENT Q U E S T I O N S Indicate whether statements 1 to 8 are true or false. 1. A structural member whose length is much greater than its cross-section dimensions is always referred to as a rod. (A) true (B) false 2. The thickness of a plate is of similar magnitude to its length and breadth. (A) true (B) false 3. Cylindrical components can be treated as membranes provided the diameter is much larger than the wall thickness. (A) true (B) false 4. There are three types of stress; direct, shear and bending stress. (A) true (B) false 5. A plate subjected to a unidirectional direct stress experiences strain only in the direction of the stress. (A) true (B) false
SUMMARY 6. The elastic stress-strain behaviour of an isotropic material can be characterized by only two elastic constants. (A) true (B) false 7. Principal stresses are the maximum and minimum shear stresses at a point in a body. (A) true (B) false 8. The failure of a stressed component can always be predicted by the von Mises criterion. (A) true (B) false For each of the statements of questions 9 to 11, one or more of the completions given are correct. Mark the correct completions. 9. Stress (A) (B) (C) (D) (E)
is the total load on a component, is a normalized measure of load, arises only from the application of a couple, always exists as two components at right angles, has units of force/length.
10. Strain (A) (B) (C) (D) (E)
is an alternative term for displacement, has units of 1/length, is a non-dimensional measure of deformation, is caused by the application of a stress, can be related to curvature of a bent beam.
11. Poisson's ratio (A) (B) (C) (D)
defines the curvature of a plate, is an alternative property to shear modulus, has the same units as stress, relates a longitudinal direct strain to the corresponding transverse strain, (E) is the same as shear strain. Each of the sentences in questions 12 to 16 consists of an assertion followed by a reason. Answer: (A) if both assertion and reason are true statements and the reason is a correct explanation of the assertion,
221
STIFFNESS, STRENGTH AND RELATED TOPICS
222 (B) if a (C) if (D) if (E) if
both assertion and reason are true statements but the reason is not correct explanation of the assertion, the assertion is true but the reason is a false statement, the assertion is false but the reason is a true statement, both the assertion and reason are false statement.
12. The loads acting on a component are sub-divided into forces and couples because forces can be sub-divided into tension and compression. 13. In a plate loaded in its plane the orthogonal direct strains are deter mined by both direct stresses because the shear strain is determined by the shear stress. 14. An isotropic material always behaves in a hnear elastic fashion shear modulus is related to Young's modulus and Poisson's ratio.
because
15. A plate loaded by couples will exhibit anticlastic curvature because direct stresses vary linearly through the thickness.
the
16. In a pressurized cylinder the axial and h o o p stresses have the same magnitude because the axial strain is greater than the circumferential strain.
ANSWERS Problems 6.1 6.2
0,011mm, (a) Yes, (b) No.
Self-assessment L B ; 2, B; 3. A; 4. B; 5, B; 6. A; 7. B; 8, B; 9. B; 10, C, D, E; 11. D; 12. B; 13, B; 14. D; 15. B; 16. E.